Simplify and expand the following expression: $ \dfrac{3}{2z - 12}- \dfrac{4}{3z + 9}+ \dfrac{4}{z^2 - 3z - 18} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{3}{2z - 12} = \dfrac{3}{2(z - 6)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{4}{3z + 9} = \dfrac{4}{3(z + 3)}$ We can factor the quadratic in the third term: $ \dfrac{4}{z^2 - 3z - 18} = \dfrac{4}{(z - 6)(z + 3)}$ Now we have: $ \dfrac{3}{2(z - 6)}- \dfrac{4}{3(z + 3)}+ \dfrac{4}{(z - 6)(z + 3)} $ The least common multiple of the denominators is: $ 6(z - 6)(z + 3)$ In order to get the first term over $6(z - 6)(z + 3)$ , multiply by $\dfrac{3(z + 3)}{3(z + 3)}$ $ \dfrac{3}{2(z - 6)} \times \dfrac{3(z + 3)}{3(z + 3)} = \dfrac{9(z + 3)}{6(z - 6)(z + 3)} $ In order to get the second term over $6(z - 6)(z + 3)$ , multiply by $\dfrac{2(z - 6)}{2(z - 6)}$ $ \dfrac{4}{3(z + 3)} \times \dfrac{2(z - 6)}{2(z - 6)} = \dfrac{8(z - 6)}{6(z - 6)(z + 3)} $ In order to get the third term over $6(z - 6)(z + 3)$ , multiply by $\dfrac{6}{6}$ $ \dfrac{4}{(z - 6)(z + 3)} \times \dfrac{6}{6} = \dfrac{24}{6(z - 6)(z + 3)} $ Now we have: $ \dfrac{9(z + 3)}{6(z - 6)(z + 3)} - \dfrac{8(z - 6)}{6(z - 6)(z + 3)} + \dfrac{24}{6(z - 6)(z + 3)} $ $ = \dfrac{ 9(z + 3) - 8(z - 6) + 24} {6(z - 6)(z + 3)} $ Expand: $ = \dfrac{9z + 27 - 8z + 48 + 24}{6z^2 - 18z - 108} $ $ = \dfrac{z + 99}{6z^2 - 18z - 108}$